Optimal. Leaf size=179 \[ -\frac{i b d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e^2}+\frac{i b d \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^2}+\frac{d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^2}+\frac{a x}{e}-\frac{b \log \left (c^2 x^2+1\right )}{2 c e}+\frac{b x \tan ^{-1}(c x)}{e} \]
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Rubi [A] time = 0.158595, antiderivative size = 179, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {4876, 4846, 260, 4856, 2402, 2315, 2447} \[ -\frac{i b d \text{PolyLog}\left (2,1-\frac{2}{1-i c x}\right )}{2 e^2}+\frac{i b d \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{2 e^2}+\frac{d \log \left (\frac{2}{1-i c x}\right ) \left (a+b \tan ^{-1}(c x)\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(1-i c x) (c d+i e)}\right )}{e^2}+\frac{a x}{e}-\frac{b \log \left (c^2 x^2+1\right )}{2 c e}+\frac{b x \tan ^{-1}(c x)}{e} \]
Antiderivative was successfully verified.
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Rule 4876
Rule 4846
Rule 260
Rule 4856
Rule 2402
Rule 2315
Rule 2447
Rubi steps
\begin{align*} \int \frac{x \left (a+b \tan ^{-1}(c x)\right )}{d+e x} \, dx &=\int \left (\frac{a+b \tan ^{-1}(c x)}{e}-\frac{d \left (a+b \tan ^{-1}(c x)\right )}{e (d+e x)}\right ) \, dx\\ &=\frac{\int \left (a+b \tan ^{-1}(c x)\right ) \, dx}{e}-\frac{d \int \frac{a+b \tan ^{-1}(c x)}{d+e x} \, dx}{e}\\ &=\frac{a x}{e}+\frac{d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}-\frac{(b c d) \int \frac{\log \left (\frac{2}{1-i c x}\right )}{1+c^2 x^2} \, dx}{e^2}+\frac{(b c d) \int \frac{\log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{1+c^2 x^2} \, dx}{e^2}+\frac{b \int \tan ^{-1}(c x) \, dx}{e}\\ &=\frac{a x}{e}+\frac{b x \tan ^{-1}(c x)}{e}+\frac{d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}+\frac{i b d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^2}-\frac{(i b d) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-i c x}\right )}{e^2}-\frac{(b c) \int \frac{x}{1+c^2 x^2} \, dx}{e}\\ &=\frac{a x}{e}+\frac{b x \tan ^{-1}(c x)}{e}+\frac{d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2}{1-i c x}\right )}{e^2}-\frac{d \left (a+b \tan ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{e^2}-\frac{b \log \left (1+c^2 x^2\right )}{2 c e}-\frac{i b d \text{Li}_2\left (1-\frac{2}{1-i c x}\right )}{2 e^2}+\frac{i b d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+i e) (1-i c x)}\right )}{2 e^2}\\ \end{align*}
Mathematica [A] time = 1.45844, size = 329, normalized size = 1.84 \[ \frac{\frac{b \left (i c d \text{PolyLog}\left (2,e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )-i c d \text{PolyLog}\left (2,-e^{2 i \tan ^{-1}(c x)}\right )+e \sqrt{\frac{c^2 d^2}{e^2}+1} \tan ^{-1}(c x)^2 e^{i \tan ^{-1}\left (\frac{c d}{e}\right )}-\frac{1}{2} \pi c d \log \left (c^2 x^2+1\right )-e \log \left (c^2 x^2+1\right )+2 i c d \tan ^{-1}(c x) \tan ^{-1}\left (\frac{c d}{e}\right )-2 c d \tan ^{-1}(c x) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )-2 c d \tan ^{-1}\left (\frac{c d}{e}\right ) \log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )}\right )+2 c d \tan ^{-1}\left (\frac{c d}{e}\right ) \log \left (\sin \left (\tan ^{-1}\left (\frac{c d}{e}\right )+\tan ^{-1}(c x)\right )\right )-i c d \tan ^{-1}(c x)^2-i \pi c d \tan ^{-1}(c x)+2 c d \tan ^{-1}(c x) \log \left (1+e^{2 i \tan ^{-1}(c x)}\right )-\pi c d \log \left (1+e^{-2 i \tan ^{-1}(c x)}\right )-e \tan ^{-1}(c x)^2+2 c e x \tan ^{-1}(c x)\right )}{c}-2 a d \log (d+e x)+2 a e x}{2 e^2} \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.045, size = 235, normalized size = 1.3 \begin{align*}{\frac{ax}{e}}-{\frac{ad\ln \left ( ecx+dc \right ) }{{e}^{2}}}+{\frac{bx\arctan \left ( cx \right ) }{e}}-{\frac{\arctan \left ( cx \right ) bd\ln \left ( ecx+dc \right ) }{{e}^{2}}}-{\frac{b\ln \left ({c}^{2}{d}^{2}-2\, \left ( ecx+dc \right ) cd+ \left ( ecx+dc \right ) ^{2}+{e}^{2} \right ) }{2\,ce}}-{\frac{{\frac{i}{2}}bd\ln \left ( ecx+dc \right ) }{{e}^{2}}\ln \left ({\frac{ie-ecx}{dc+ie}} \right ) }+{\frac{{\frac{i}{2}}bd\ln \left ( ecx+dc \right ) }{{e}^{2}}\ln \left ({\frac{ie+ecx}{ie-dc}} \right ) }-{\frac{{\frac{i}{2}}bd}{{e}^{2}}{\it dilog} \left ({\frac{ie-ecx}{dc+ie}} \right ) }+{\frac{{\frac{i}{2}}bd}{{e}^{2}}{\it dilog} \left ({\frac{ie+ecx}{ie-dc}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} a{\left (\frac{x}{e} - \frac{d \log \left (e x + d\right )}{e^{2}}\right )} + 2 \, b \int \frac{x \arctan \left (c x\right )}{2 \,{\left (e x + d\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x \arctan \left (c x\right ) + a x}{e x + d}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b \operatorname{atan}{\left (c x \right )}\right )}{d + e x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )} x}{e x + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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